Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the energy difference between the two lowest energy levels for a proton in a cubical box with side length 1.00 \(\times\) 10\(^{-14}\) m, the approximate diameter of a nucleus?

Short Answer

Expert verified
The energy difference \(\Delta E = 2.42 \times 10^{-13}\) J.

Step by step solution

01

Understand the system

The problem involves a proton confined in a cubical box, a situation where quantum mechanics is applicable. We will use the particle in a box model to calculate energy levels.
02

Know the Energy Formula for a Cubical Box

The energy levels for a particle in a 3D box are given by\[E_{n_x, n_y, n_z} = \frac{h^2}{8mL^2} (n_x^2 + n_y^2 + n_z^2)\]where \(h\) is Planck's constant, \(m\) is the mass of the proton, \(L\) is the side of the box, and \(n_x, n_y, n_z\) are quantum numbers (1, 2, 3,...).
03

Set Quantum Numbers for Two Lowest Energy Levels

The lowest energy state is the ground state with \(n_x = n_y = n_z = 1\). The next energy state involves exciting one quantum number by an integer; often it's \(n_x = 2, n_y = n_z = 1\).
04

Calculate Ground State Energy

For the ground state, where \(n_x = n_y = n_z = 1\), the energy is:\[E_{1,1,1} = \frac{h^2}{8mL^2}(1^2 + 1^2 + 1^2) = \frac{3h^2}{8mL^2}\]
05

Calculate First Excited State Energy

For the first excited state, a potential scenario is \(n_x = 2, n_y = n_z = 1\), giving:\[E_{2,1,1} = \frac{h^2}{8mL^2}(2^2 + 1^2 + 1^2) = \frac{6h^2}{8mL^2}\]
06

Compute Energy Difference

The energy difference between the first excited state and the ground state is:\[\Delta E = E_{2,1,1} - E_{1,1,1} = \left(\frac{6h^2}{8mL^2}\right) - \left(\frac{3h^2}{8mL^2}\right) = \frac{3h^2}{8mL^2}\]
07

Insert Values to Find Numerical Result

Insert the values for Planck's constant \(h = 6.626 \times 10^{-34}\, \text{Js}\), the mass of the proton \(m = 1.673 \times 10^{-27}\, \text{kg}\), and the side length \(L = 1.00 \times 10^{-14}\, \text{m}\):\[\Delta E = \frac{3 \times (6.626 \times 10^{-34})^2}{8 \times 1.673 \times 10^{-27} \times (1.00 \times 10^{-14})^2}\]
08

Calculate Final Numerical Solution

Carry out the computation in step 7 to find \(\Delta E\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Particle in a Box
The "particle in a box" is a popular model in quantum mechanics. It simplifies the complex nature of particles confined in a limited space, like the nucleus of an atom. Think of it as a way to visualize how particles behave when trapped in a very tiny space.

In this model, the box is usually considered to have infinitely high walls, meaning the particle cannot escape. It's a useful concept because it allows us to calculate the possible energy levels of the particle based on their quantized nature. In our specific problem, a proton is considered to be in such a box, which represents the confined space of a nucleus. The cubical shape we use is idealized to make the math easier, but it captures the essential physics.
Energy Levels
Energy levels in quantum mechanics are like "rungs on a ladder." A particle can only exist on these specific steps and nowhere in between.

When we talk about energy levels in the "particle in a box" model, these levels are determined numerically using an equation. This equation includes quantum numbers, Planck's constant, the mass of the particle, and the size of the confinement region (like our cubical box). Different arrangements of quantum numbers create different energy levels.
  • The ground state is the lowest energy level, where the quantum numbers are all minimized (e.g., 1, 1, 1).
  • Excited states are higher energy levels where at least one quantum number is increased.
This is how we derive energy levels for our proton in the box.
Planck's Constant
Planck's constant is a fundamental constant in physics that plays an essential role in quantum mechanics. It defines the size of the quantization found in nature.

In the formula for the "particle in a box," Planck's constant appears as a squared value. This constant, denoted as \( h \), is incredibly small, approximately \( 6.626 \times 10^{-34} \text{Js} \). Its tiny size indicates that the energy differences we're discussing are extremely small at the quantum level.

Planck's constant essentially "scales" the quantum mechanical world to its microscopic nature, affecting how we calculate energy levels and the precise nature of quantum jumps between these levels.
Quantum Numbers
Quantum numbers are key to understanding the quantized nature of particles in quantum mechanics. In the "particle in a box" model, they help define the possible energy levels.

There are typically three quantum numbers \( n_x \), \( n_y \), and \( n_z \) in a three-dimensional box. Each of these numbers can be an integer starting from 1.
  • The smallest numbers represent the lowest energy state, or ground state.
  • Higher numbers indicate excited states.
These values are crucial for calculating the energy levels using the formula provided. By changing these numbers, you compute different energy scenarios for the particle, such as moving from ground state to an excited state, which is what causes energy differences as per our calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A hydrogen atom in the \(n = 1\), \(ms = - {1\over2}\) state is placed in a magnetic field with a magnitude of 1.60 \(T\) in the \(+z\)- direction. (a) Find the magnetic interaction energy (in electron volts) of the electron with the field. (b) Is there any orbital magnetic dipole moment interaction for this state? Explain. Can there be an orbital magnetic dipole moment interaction for \(n\neq 1\)?

Spectral Analysis. While studying the spectrum of a gas cloud in space, an astronomer magnifies a spectral line that results from a transition from a \(p\) state to an \(s\) state. She finds that the line at 575.050 nm has actually split into three lines, with adjacent lines 0.0462 nm apart, indicating that the gas is in an external magnetic field. (Ignore effects due to electron spin.) What is the strength of the external magnetic field?

You are studying the absorption of electromagnetic radiation by electrons in a crystal structure. The situation is well described by an electron in a cubical box of side length \(L\). The electron is initially in the ground state. (a) You observe that the longest-wavelength photon that is absorbed has a wavelength in air of \(\lambda\) = 624 nm. What is \(L\)? (b) You find that \(\lambda\) = 234 nm is also absorbed when the initial state is still the ground state. What is the value of \(n$$^2\) for the final state in the transition for which this wavelength is absorbed, where \(n$$^2\) = \(n$$_X^2\) + \(n$$_y^2\) + \(n$$_z^2\) ? What is the degeneracy of this energy level (including the degeneracy due to electron spin)?

In the Bohr model, what is the principal quantum number \(n\) at which the excited electron is at a radius of 1 \(\mu\)m? (a) 140; (b) 400; (c) 20; (d) 81.

A hydrogen atom in a 3\(p\) state is placed in a uniform external magnetic field \(\vec B\). Consider the interaction of the magnetic field with the atom's orbital magnetic dipole moment. (a) What field magnitude \(B\) is required to split the 3\(p\) state into multiple levels with an energy difference of 2.71 \(\times\) 10\(^{-5}\) eV between adjacent levels? (b) How many levels will there be?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free